Problem: Compute \[\dfrac{2^3-1}{2^3+1}\cdot\dfrac{3^3-1}{3^3+1}\cdot\dfrac{4^3-1}{4^3+1}\cdot\dfrac{5^3-1}{5^3+1}\cdot\dfrac{6^3-1}{6^3+1}.\]
Solution: Let $S$ be the value of the given expression. Using sum and difference of cubes to factor, we get \[\begin{aligned} S &= \dfrac{(2-1)(2^2+2+1)}{(2+1)(2^2-2+1)}\cdot\dfrac{(3-1)(3^2+3+1)}{(3+1)(3^2-3+1)} \cdot\dfrac{(4-1)(4^2+4+1)}{(4+1)(4^2-4+1)}\cdot\dfrac{(5-1)(5^2+5+1)}{(5+1)(5^2-5+1)}\cdot\dfrac{(6-1)(6^2+6+1)}{(6+1)(6^2-6+1)} \\ &= \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{4}{6} \cdot \frac{5}{7} \cdot \frac{2^2+2+1}{2^2-2+1} \cdot \frac{3^2+3+1}{3^2-3+1} \cdot \frac{4^2+4+1}{4^2-4+1} \cdot \frac{5^2+5+1}{5^2-5+1} \cdot \frac{6^2+6+1}{6^2-6+1}.\end{aligned}\]The first product telescopes to $\tfrac{1 \cdot 2}{6 \cdot 7} = \tfrac{1}{21}$. The second product also telescopes due to the identity \[x^2 + x + 1 = (x+1)^2 - (x+1) + 1.\]That is, the terms $2^2+2+1$ and $3^2-3+1$ cancel, as do the terms $3^2+3+1$ and $4^2-4+1$, and so on, leaving just $\tfrac{6^2+6+1}{2^2-2+1} = \tfrac{43}{3}$. Thus, \[S = \frac{1}{21} \cdot \frac{43}{3} = \boxed{\frac{43}{63}}.\]